Question

Find a focus for this conic section. 9x^2+25y^2-200y+175=0

Answer 1

First of all we have to figure out what type of a conic this is.  We know it's not a parabola because it has both an x-squared term and a y-squared term.  There's a plus sign separating the squared terms so we know it also cannot be a hyperbola.  It's either a circle or an ellipse.  If this was simply a circle, though, we would not have leading coefficients on the squared terms (other than a 1).  Circles have a standard form of $(x-h)^2+(y-k)^2=r^2$.  That makes this an ellipse. Let's group together the x terms and the y terms and move the constant over and complete the squares to see what we have.  $9x^2+25y^2-200y=-175$.  Since there's only 1 term with the x squared expression we cannot complete the square on the x's but we can on the y terms.  First, though, the rule for completing the square is that the leading coefficient has to be a 1 and ours is a 25, so let's factor it out. $9x^2+25(y^2-8y)=-175$.  To complete the square we take half the linear term, square it, and add it to both sides.  Our linear term is 8.  Half of 8 is 4, and 4 squared is 16. So we add a 16 into the parenthesis, BUT we cannot disregard that 26 sitting out front there.  It refuses to be ignored.  It is still considered a multiplier.  So what we are really adding on to the right side is 25*16 which is 400.  $9x^2+25(y^2-8y+16)=-175+400$ which simplifies to $9x^2+25(y^2-8y+16)=225$.  The standard form of an ellipse is $\frac{(x-h)^2}{a^2}+ \frac{(y-k)^2}{b^2} =1$ if its horizontal axis is the transverse axis, or $\frac{(x-h)^2}{b^2}+ \frac{(y-k)^2}{a^2}=1$ if its vertical axis is the transverse axis.  Notice that the a and b moved as the difference between the 2 equations.  A is always the larger value and dictates which ellipse you have, horizontal or vertical.  Our equation has a 225 on the right, so we will divide both sides by 225 to get that much-needed 1 on the right: $\frac{x^2}{25}+ \frac{(y-4)^2}{9}=1$.  Because 25 is larger than 9, this is a horizontal ellipse, our a value is the square root of 25 which is 5, and our b value is the square root of 9 which is 3.  The center is (0, 4).  You want the focus and now we can find it.  The formula for the focus is $c^2=a^2-b^2$ where c is the distance from the center to the focus.  We have an a and a b to find c: $c^2=25-9$, which gives us that c=4.  The focus is 4 units from the center and always lies on the transverse axis.  It shares a y value with the center and moves from the x-coordinate of the center the value of c.  Our center is (0, 4) so our y value of the focus is 4; our x coordinate of the center is 0, so the x value of the focus is 4.  The coordinates of the focus are (4, 4).